Proof of $\lim_{h\to 0}\frac{t^h-1}{h}=\ln t$ please

Question

I should know this, but I don't. Please help me understand why $$\lim_{h\to 0}\frac{t^h-1}{h}=\ln t$$

Answer 1

I thought it might be instructive to present an approach that relies on an elementary analysis. Here we use only a well-known pair of inequalities and the squeeze theorem.

Note that $t^h=e^{h\log(t)}$. Now use the inequalities

$$1+x\le e^x \le \frac1{1-x}$$

for $x<1$.

Finish by using the squeeze theorem.

Answer 2

Notice that

$$ \lim_{h \to 0} \frac{ t^h - t^0}{h-0} = \frac{d}{d h} t^h \bigg|_{h \to 0 } = t^h \ln t \bigg|_{h=0} = \ln t$$

Answer 3

we have the problem: $$L=\lim_{h\to 0}\frac{x^h-1}{h}$$ firstly notice that: $$\lim_{h\to 0}\left[x^h-1\right]=\lim_{h\to 0}\left[h\right]=0$$ so this satisfies the conditions for L'Hopitals rule which states: $$\lim_{y\to 0}\frac{f(y)}{g(y)}=\lim_{y\to 0}\frac{f'(y)}{g'(y)}$$ and so: $$L=\lim_{h\to 0}\frac{x^h\ln(x)}{1}=x^0\ln(x)=\ln(x)$$ and so there is the proof